i. For a cube of resistors each of resistance $R$ the equivalent resistance between two edges on the same face is ${\mathrm{R}}_{eq}=\frac{5\mathrm{R}}{9}$

ii. If $AB$ is removed, assuming the equivalent resistance between remaining resistors is $x,$

$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{x}$

$R_{eq}=\frac{Rx}{R+x}$

$\therefore \frac{5}{9}=\frac{1\left(x\right)}{1+x}=\frac{x}{1+x} ....\left(\because R=1\Omega \right)$

$\therefore 5(1+x)=9x$

$\therefore 5+5x=9x$

$\therefore 5=4x$

$\therefore x=\frac{5}{4}\Omega$

$\therefore x=1.25\Omega$